Confidence Intervals. In Exercises 9–24, construct the confidence interval estimate of the mean. Arsenic in Rice Listed below are amounts of arsenic (μg, or micrograms, per serving) in samples of brown rice from California (based on data from the Food and Drug Administration). Use a 90% confidence level. The Food and Drug Administration also measured amounts of arsenic in samples of brown rice from Arkansas. Can the confidence interval be used to describe arsenic levels in Arkansas? 5.4 5.6 8.4 7.3 4.5 7.5 1.5 5.5 9.1 8.7

Ten amounts of levels of arsenic in brown rice are given. The confidence level for this sample is 90%.

Confidence interval is a range of values that is used to estimate the true value of population parameters. The formula for the confidence interval is stated below.

$\overline{x}-E<\mu <\overline{x}+E$

Here,$\overline{x}$is a sample mean, and$E$ is the margin of error.

When the sample is large, the formula for the confidence interval is

$\overline{x}-Z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n}}<\mu <\overline{x}+Z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n}}$

When the sample is less than 30, the formula is changed with t- distribution, as follows.

$\overline{x}-t_{\frac{\alpha }{2}}\times \frac{s}{\sqrt{n}}<\mu <\overline{x}+t_{\frac{\alpha }{2}}\times \frac{s}{\sqrt{n}}$

Let $\text{n}=10$be the amounts of levels of arsenic in rice.

The mean value is given by,

$$\begin{gathered} \overline{x}=\frac{{\sum }_{i=1}^n{x}_i}{n} \\ =\frac{63.5}{10} \\ =6.35 \end{gathered}$$

The mean of the given sample is $6.35$.

The standard deviation of the sample is given below.

$$\begin{gathered} s=\sqrt{\frac{{\sum }_{i=1}^n{\left(x_i-\overline{x}\right)}^2}{n-1}} \\ =\sqrt{\frac{{\left(5.4-6.35\right)}^2+{\left(5.6-6.35\right)}^2+\cdots +{\left(8.7-6.35\right)}^2}{10-1}} \\ =\sqrt{\frac{48.645}{9}} \\ =2.3249 \end{gathered}$$

The standard deviation of the sample is$2.3249$.

Assume that the population is normally distributed, and the samples are randomly selected. As the population standard deviation is unknown, t-distribution is used.

The degree of freedom is

$$\begin{gathered} \text{df}=n-1 \\ =10-1 \\ =9 \end{gathered}$$

At $90\%$confidence, $\alpha =0.10$.

The critical value is obtained from the t-table corresponding to the row with a value of 9 and column 0.10 (two-tailed).

$$\begin{gathered} t_{\text{crit}}=t_{\frac{\alpha }{2}} \\ =t_{\frac{0.10}{2}} \\ =1.833 \end{gathered}$$

The formula for the margin of error is stated below.

$E=t_{crit}\times \frac{s}{\sqrt{n}}$

Substitute the calculated values in the formula. So,

$$\begin{gathered} E=1.833\times \frac{2.32}{\sqrt{10}} \\ =1.3477 \end{gathered}$$

Thus, the margin of error is 1.3477.

Substitute all the calculated values in the formula.

$$\begin{gathered} \text{Confidence}\text{interval}=\left(6.35-1.344\right)<\mu <\left(6.35+1.344\right) \\ =5.002<\mu <7.698 \\ =\left(5.00,7.70\right) \end{gathered}$$

Thus, the 90% confidence level for the mean amount of arsenic in brown rice is $\left(5.00\mu g,7.70\mu g\right)$.

The confidence level is computed from the sampled brown rice from California for the arsenic level. Thus, it provides an estimate for the population of rice in California. It will not be helpful to describe the arsenic levels in brown rice in Arkansas.